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Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem number 2 from the American Invitational Mathematics Examination I, AIME I, 2012 based on Arithmetic Sequence Problem.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to \(715\). The first term of the sequence is increased by \(1\), the second term is increased by \(3\), the third term is increased by \(5\), and in general, the \(k\)th term is increased by the \(k\)th odd positive integer. The terms of the new sequence add to \(836\). Find the sum of the first, last, and middle terms of the original sequence.

  • is 107
  • is 195
  • is 840
  • cannot be determined from the given information

Key Concepts


Number Theory


Check the Answer

But try the problem first…

Answer: is 195.

Suggested Reading

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

Try with Hints

First hint

After the adding of the odd numbers, the total of the sequence increases by \(836 – 715 = 121 = 11^2\).

Second Hint

Since the sum of the first \(n\) positive odd numbers is \(n^2\), there must be \(11\) terms in the sequence, so the mean of the sequence is \(\frac{715}{11} = 65\).

Final Step

Since the first, last, and middle terms are centered around the mean, then \(65 \times 3 = 195\)

Hence option B correct.

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